Question

Hazel has an assortment of red, blue, and green balls. The number of red balls is 2/3 the number of blue balls. The number of green balls is 1 more than 1/3 the number of blue balls. In total, she has 15 balls.
An equation created to find the number of blue balls will have

- no solution
- one solution
- infinitely many solutions

Answer 1

Answer:

x = 4$\frac{2}{3}$      y = 7        and    z =3$\frac{1}{3}$

This implies the equation has just one solution

Step-by-step explanation:

To create the equation, we need to be able to write the information  or interpret the question mathematically.

Let x equal to the number of red balls.

Let y equal to the number of blue balls.

Let z equal to the number of green balls.

From the question; "The number of red balls is 2/3 the number of blue balls" can be mathematically written as :  x = $\frac{2}{3}$ y  ---------------------------(1)

The next statement; "The number of green balls is 1 more than 1/3 the number of blue balls" can be written mathematically as: z = 1+$\frac{1}{3}$ y ----------------------------(2)

The next statement; "she has 15 balls."  can be mathematically written as:     x + y + z = 15 ----------------------------------------(3)

Substitute  equation (1) and equation (2) into equation (3)

$\frac{2}{3}$ y + y +1 +$\frac{1}{3}$ y  =  15

We can rearrange this equation

$\frac{2}{3}$ y  +$\frac{1}{3}$ y + y +1 =  15

$\frac{3}{3}$ y  + y + 1 = 15

y + y + 1 = 15

2y + 1 = 15

subtract 1 from both-side of the equation

2y + 1 -1 = 15 -1

2y = 14

Divide both-side of the equation by 2

2y/2 = 14/2

y = 7

Substitute y = 7 into equation (1)

x = $\frac{2}{3}$ y

x = $\frac{2}{3}$ (7)

x = 14/3

x = 4$\frac{2}{3}$

Substitute y= 7 in equation (2)

z = 1+$\frac{1}{3}$ y

z = 1+$\frac{1}{3}$ (7)

z = 1+ 7/3

z = 10/3

z =3$\frac{1}{3}$

Therefore;

x = 4$\frac{2}{3}$      y = 7        and    z =3$\frac{1}{3}$

This implies the equation has just one solution.

Answer 2

It would have one solution since you know the total number of balls